$x > 4$
$\rightarrow$ $x - 1 > 3$
$\rightarrow$ $(x - 1)^2 > 9$ and obviously if $(x - 1)^2 > 9$ then $x^2 > 9$
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See: https://math.stackexchange.com/questions/568780/why-cant-you-square-both-sides-of-an-equation – NoChance Nov 18 '18 at 04:50
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It's not true that $(x-1)^{2}>9$ implies that $x^{2}>9$. For example, if $x=-2.5$, then $(x-1)^{2}>9$ but $x^{2}<9$. – Brian Borchers Nov 18 '18 at 04:52
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@BrianBorchers $x>4$ from the start. – Rócherz Nov 18 '18 at 04:53
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1I guess somewhere you need to use that $x>0$ in an essential way. – Andres Mejia Nov 18 '18 at 04:53
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Why don't you use $$x>4\implies x^2>16>9$$
Note: How did we arrive at the second step from the first one?
Soham
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Oh that's probably better haha. But does mine make sense at least? Like if I used it on an exam would I get full marks? – ming Nov 18 '18 at 05:30
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1@ming Your argument is also correct however to be on the safe side in an exam, you should write that "...since,we know, $x^2$ is an increasing function, hence the result follows...". You should explicitly mention that $x^2$ is an increasing function. – Soham Nov 18 '18 at 15:50
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There are two underlying facts in what you label as "obvious" in your last step.
$x-1>3 \implies x-1>0$.
$x>x-1$ (this one is universal) and $x-1>0 \implies x^2>(x-1)^2$.
But of course the fastest way is as Andrés Mejía and tatan suggested.
Rócherz
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